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href="/2022/06/04/2022%E5%B9%B4%E5%B1%B1%E4%B8%9C%E7%9C%81ICPC%E7%9C%81%E8%B5%9B%E6%B8%B8%E8%AE%B0/" title="2022年山东省ICPC省赛游记">     <img class="post_bg" src="https://pic.imgdb.cn/item/629b123b09475431292426f7.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="2022年山东省ICPC省赛游记"></a></div><div class="recent-post-info"><a class="article-title" href="/2022/06/04/2022%E5%B9%B4%E5%B1%B1%E4%B8%9C%E7%9C%81ICPC%E7%9C%81%E8%B5%9B%E6%B8%B8%E8%AE%B0/" title="2022年山东省ICPC省赛游记">2022年山东省ICPC省赛游记</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-06-04T07:59:36.000Z" title="发表于 2022-06-04 15:59:36">2022-06-04</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-06-04T08:38:39.956Z" title="更新于 2022-06-04 16:38:39">2022-06-04</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E5%8F%82%E8%B5%9B%E8%AE%B0%E5%BD%95/">参赛记录</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E8%AE%B2%E6%95%85%E4%BA%8B/">讲故事</a></span></div><div class="content">由于比完赛之后有一大坨考试要考（选修课和必修课我是一课也没敢学啊），所以等到突击复习而且考完比较集中的几门考试之后再来记录一下比赛过程。（也不知道计网能不能过）
关于比赛过程Day1 热身赛题目很水，就是三人的英语水平实在是有些许拉跨，搞得读C的时候慢了不少，手速rank15，还是很开心的。

Day2 正式赛因为我们当天要全员核酸，所以身份核验完之后我们就润去做核酸了，在进去选排队的核酸口的时候选了人很多的一队，当时就感觉不妙（害怕比赛开题也选错了题）。
做完回去路上口嗨，wushi（十分强大的大一学弟）还是十分的紧张，开始前几分钟的过题人数并不多，没有比较明显的签到题，就只有A题过的比较多，我们去看了A，发现感觉会是一个十分恶心的题（事实也是如此），所以打算先放一放去找真正的签到题。这时候wushi把H丢给了我说这是一个铁签，快速给我喂了题意发现雀食是个签到题，随便写了一发交上去1A，一看榜单竟是FB（竞赛生涯中的首个一血）并且持续了比较长的时间只有我们一个队伍过（骄傲）。
调整了一波心态去做A，之后噩梦开始了，看了十几二十分钟发现了正确的tick，码完之后狂wa，其实是因为没有特 ...</div></div></div><div class="recent-post-item"><div class="post_cover right_radius"><a href="/2022/04/11/%E5%B1%B1%E4%B8%9C%E5%A4%9A%E6%A0%A1%E9%9B%86%E8%AE%AD1%E6%80%BB%E7%BB%93%EF%BC%88%E6%B5%81%E6%B0%B4%E8%B4%A6%EF%BC%89%E9%99%84%E4%B8%8D%E5%AE%8C%E5%85%A8%E9%A2%98%E8%A7%A3/" title="山东多校集训1总结（流水账）附不完全题解">     <img class="post_bg" src="https://pic.imgdb.cn/item/6253b949239250f7c54a1415.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="山东多校集训1总结（流水账）附不完全题解"></a></div><div class="recent-post-info"><a class="article-title" href="/2022/04/11/%E5%B1%B1%E4%B8%9C%E5%A4%9A%E6%A0%A1%E9%9B%86%E8%AE%AD1%E6%80%BB%E7%BB%93%EF%BC%88%E6%B5%81%E6%B0%B4%E8%B4%A6%EF%BC%89%E9%99%84%E4%B8%8D%E5%AE%8C%E5%85%A8%E9%A2%98%E8%A7%A3/" title="山东多校集训1总结（流水账）附不完全题解">山东多校集训1总结（流水账）附不完全题解</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-04-11T04:18:00.000Z" title="发表于 2022-04-11 12:18:00">2022-04-11</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-04-11T07:12:06.776Z" title="更新于 2022-04-11 15:12:06">2022-04-11</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E5%8F%82%E8%B5%9B%E8%AE%B0%E5%BD%95/">参赛记录</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E8%AE%B2%E6%95%85%E4%BA%8B/">讲故事</a></span></div><div class="content">关于比赛过程吃完饭去开门，一开门进去发现导员在睡觉（疫情期间导员都留校住宿了），之后说明了一下去楼上打印试题，打了一份半木的纸了，问题不大……
比赛开始前六分钟我们在激情讨论前一天蓝桥杯的一些题目（大雾，突然发现开始了直接点开榜单疯狂刷新，之后发现有人过了A，我和wushi读A秒出思路，但是由于我写的时候脑子不太灵光，写出了几个锅wa了几发，途中发现E是个水题浅wa一发就过了，M也是个水题，讨论了一下就过了。
下一个看B，和wushi想了个状态机dp，之后写完测完样例直接交了，还在跑的时候gyx说样例解释里说可以是浮点数（说晚了已经交了，改了浮点数又wa了一发，仔细一想发现炸double了，wushi直接一手玄学大法当数比较大时都除1e6过掉了，赛后发现是贪心（我们看的atcoder官方题解里用的异或？。
接下来有两个选择K和H，我们是都读了，之后发现K是个博弈论，本着博弈论要不根本不会，要不就是伞兵题的原则，我们尝试起了这个题，涂涂画画十分钟发现是个诈骗题，甚至wa了一发特判。
H题看了一下数据范围一眼dp，wushi给出了状态表示和状态转移方程，没有发现第三维开小了疯狂wa，好在最 ...</div></div></div><div class="recent-post-item"><div class="post_cover left_radius"><a href="/2022/03/21/%E6%AF%8F%E6%97%A5%E4%B8%80DP-6-Leetcode-2172-%E7%8A%B6%E5%8E%8BDP/" title="每日一DP(6) Leetcode 2172(状压DP)">     <img class="post_bg" src="https://pic.imgdb.cn/item/6237ef7527f86abb2af1c508.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="每日一DP(6) Leetcode 2172(状压DP)"></a></div><div class="recent-post-info"><a class="article-title" href="/2022/03/21/%E6%AF%8F%E6%97%A5%E4%B8%80DP-6-Leetcode-2172-%E7%8A%B6%E5%8E%8BDP/" title="每日一DP(6) Leetcode 2172(状压DP)">每日一DP(6) Leetcode 2172(状压DP)</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-03-21T02:40:26.000Z" title="发表于 2022-03-21 10:40:26">2022-03-21</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-03-21T03:22:42.481Z" title="更新于 2022-03-21 11:22:42">2022-03-21</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/">每日一题</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/">动态规划</a></span></div><div class="content">昨天排位赛遇到的一个题，由于自己基本没怎么写过状压DP所以寄了。
题目原题链接：2172. 数组的最大与和
题目大意有$x$个篮子，每个篮子都有编号，还有一个数组$a$，每个篮子至多装$2$个数组中的数，求每个数与它所在的篮子编号的按位与运算之和。
解法一：二进制状压DP思路数据范围很小，考虑状压DP。
因为每个篮子可以放两个数，我们可以把$x$个篮子看作$x*2$个篮子，每个篮子只能装一个数，并且篮子的编号为$\frac{i}{2}+1$。
我们可以用二进制数字$x$来表示篮子中是否存放数字的方案，从低到高的第$i$位如果为$1$，则表示这个篮子中放了数字，否则表示篮子为空。
设$x$的二进制表示中，有$c$个$1$，定义$dp[x]$表示将$a$中的前$c$个数字放到篮子中，并且方案为$x$时的最大按位与和，$dp[0]=0$。
可以考虑将$a[c]$放到一个空蓝子时的状态转移方程（由于下标从$0$开始，所以$a[c]$还未被放到篮子中），则可以枚举空蓝子的位置$j$，则改篮子对应的位置为$\frac{j}{2}+1$，则：

dp[i+2^j]=max(dp[i+2^j],dp[ ...</div></div></div><div class="recent-post-item"><div class="post_cover right_radius"><a href="/2022/03/18/%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0-8-%E2%80%94%E2%80%94%E8%83%8C%E5%8C%85DP/" title="算法学习笔记(8)——背包DP">     <img class="post_bg" src="https://pic.imgdb.cn/item/623403185baa1a80abe33090.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="算法学习笔记(8)——背包DP"></a></div><div class="recent-post-info"><a class="article-title" href="/2022/03/18/%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0-8-%E2%80%94%E2%80%94%E8%83%8C%E5%8C%85DP/" title="算法学习笔记(8)——背包DP">算法学习笔记(8)——背包DP</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-03-18T00:58:52.000Z" title="发表于 2022-03-18 08:58:52">2022-03-18</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-03-18T03:57:24.535Z" title="更新于 2022-03-18 11:57:24">2022-03-18</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%AE%97%E6%B3%95%E7%AC%94%E8%AE%B0/">算法笔记</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/">动态规划</a></span></div><div class="content">背包问题背包问题是一个典型的组合问题，目标是让我们在容量为$W$的背包中尽可能装价值高的物品，其中每件物品都对应自己的重量$w_i$和价值$v_i$，并且其选取方式遵循某种规则。
如果采用暴力枚举的方式，时间复杂度高达$O(2^n)$，这种复杂度很明显无法让人接受。
0-1背包问题
有$n$个物品和一个容量为$W$的背包，每个物品有重量$w_i$和价值$v_i$两种属性，要求选若干物品放入背包使背包中物品的总价值最大且背包中物品的总重量不超过背包的容量，每个物品只能选一次。

对于上述问题，每个物品只有两种状态——取或者不取（对应1和0），所以这种问题被称为0-1背包问题。
对于DP问题来说，状态表示和状态转移是最重要的两点。
对于0-1背包问题来说，使用$dp[i] [j]$来表示取到第$i$个物品时，容量为$j$的背包能够到达的最大价值。
很明显，对于第$i$个物品来说，我们只需要考虑该物品选或者不选的状态就好。
当第$i$个物品不选时，$dp[i] [j]=dp[i-1] [j]$。
而当选择第$i$个物品时，因为目前的总容量为$j$，并且需要放入第$i$个物品，则对于第$i-1 ...</div></div></div><div class="recent-post-item"><div class="post_cover left_radius"><a href="/2022/01/13/%E6%AF%8F%E6%97%A5%E4%B8%80DP-5-CF1625C-%E7%BA%BF%E6%80%A7DP/" title="每日一DP(5) CF1625C (线性DP)">     <img class="post_bg" src="https://pic.imgdb.cn/item/61dffe022ab3f51d91788e88.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="每日一DP(5) CF1625C (线性DP)"></a></div><div class="recent-post-info"><a class="article-title" href="/2022/01/13/%E6%AF%8F%E6%97%A5%E4%B8%80DP-5-CF1625C-%E7%BA%BF%E6%80%A7DP/" title="每日一DP(5) CF1625C (线性DP)">每日一DP(5) CF1625C (线性DP)</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-01-13T10:03:53.000Z" title="发表于 2022-01-13 18:03:53">2022-01-13</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-01-13T10:25:17.320Z" title="更新于 2022-01-13 18:25:17">2022-01-13</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/">每日一题</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/">动态规划</a></span></div><div class="content">真的是好久没有敲代码了，都生疏了。
而且这场的题目真的都是又臭又长，在加上我阅读水平也不高，真的是受苦场。
题目原题链接：Problem - C - Codeforces
题目大意就是说在一条长$l$的公路上，有$n$个速度牌，上面的数字代表接下来的一公里要走几分钟，并且这个速度牌在下一个速度牌出现前都生效，现在可以拔掉至多$k$个速度牌子，并且第一个牌子不能拔掉，问最短可以多长世界走完公路。
思路我一开始的思路就是硬贪，把每个中间的差值都放到一个堆里，不断的取，最后发现这很很多的问题，可能可以实现，但是异常的麻烦。而且很长世界的脱离编程让我没往DP上想，麻了麻了。
先看一下数据范围，$l$和$a_i$的范围保证了数据范围在$int$之内，再看$n$的范围：$1 \leq n \leq 500$。
很小的一个范围，考虑$O(n^3)$的算法，很容易想到DP(可惜对赛时的我不容易)。
我们在考虑转移的时候需要找一个末状态作为参照，所有考虑将$dp[i] [j]$表示为从第$i$个速度牌开始走，拔掉$j$个速度牌并且不拔第$i$个时，走完剩下路程的最小时间。
之后枚举到下一个未被拔走的速度 ...</div></div></div><div class="recent-post-item"><div class="post_cover right_radius"><a href="/2022/01/11/%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0-7-%E2%80%94%E2%80%94%E6%A0%91%E7%8A%B6%E6%95%B0%E7%BB%84/" title="算法学习笔记(7)——树状数组">     <img class="post_bg" src="https://pic.imgdb.cn/item/61dd7c6b2ab3f51d91b9e054.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="算法学习笔记(7)——树状数组"></a></div><div class="recent-post-info"><a class="article-title" href="/2022/01/11/%E7%AE%97%E6%B3%95%E5%AD%A6%E4%B9%A0%E7%AC%94%E8%AE%B0-7-%E2%80%94%E2%80%94%E6%A0%91%E7%8A%B6%E6%95%B0%E7%BB%84/" title="算法学习笔记(7)——树状数组">算法学习笔记(7)——树状数组</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2022-01-11T09:58:39.000Z" title="发表于 2022-01-11 17:58:39">2022-01-11</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2022-01-11T13:04:44.786Z" title="更新于 2022-01-11 21:04:44">2022-01-11</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%AE%97%E6%B3%95%E7%AC%94%E8%AE%B0/">算法笔记</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84/">数据结构</a></span></div><div class="content">关于树状数组我想快速的去修改某一点的值，并且不需要大量的区间查询，我们可以直接使用数组。
而需要大量区间查询不需要频繁的修改时，我们经常使用前缀和来解决。
但是上述两种方法在面对同时具有大量区间查询和单点修改需求的题目时显得比较乏力。
所有我们会引入一个新的数据结构：树状数组（Binary Index Tree, BIT）,它支持上述两种操作，并且时间复杂度都是$O(logn)$。
树状数组的实现我们可以受一些前缀和的启发，储存前缀和的数组中第$i$个元素其实是维护着从$a_1~a_i$的和，这样我们可以快速的进行区间查询。
同样的，普通数组中的第$i$个元素就相当于维护着从$a_i~a_i$的和，可以时间快速的单点修改。
因此，我们可以思考一种存储方式，也是维护某段区间，来达到同时实现不那么慢的进行区间查询已经不那么慢的进行单点修改。
树状数组就是这样的一种思路，在此之上还运用了二进制的一些思想。
如果我们想求前5项的和，则可以：

((000)_2,(101)_2]=((100)_2,(101)_2]+((000)_2,(100)_2]可以观察到每一次的加和都是将二进制表示中的最右 ...</div></div></div><div class="recent-post-item"><div class="post_cover left_radius"><a href="/2021/11/16/%E6%AF%8F%E6%97%A5%E4%B8%80DP-4-CF1061C-%E6%95%B0%E5%AD%A6-%E7%BA%BF%E6%80%A7DP-%E9%A2%84%E5%A4%84%E7%90%86/" title="每日一DP(4) CF1061C(数学+线性DP+预处理)">     <img class="post_bg" src="https://pic.imgdb.cn/item/6193c6152ab3f51d91b888ad.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="每日一DP(4) CF1061C(数学+线性DP+预处理)"></a></div><div class="recent-post-info"><a class="article-title" href="/2021/11/16/%E6%AF%8F%E6%97%A5%E4%B8%80DP-4-CF1061C-%E6%95%B0%E5%AD%A6-%E7%BA%BF%E6%80%A7DP-%E9%A2%84%E5%A4%84%E7%90%86/" title="每日一DP(4) CF1061C(数学+线性DP+预处理)">每日一DP(4) CF1061C(数学+线性DP+预处理)</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2021-11-16T14:52:56.000Z" title="发表于 2021-11-16 22:52:56">2021-11-16</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-11-16T15:48:25.169Z" title="更新于 2021-11-16 23:48:25">2021-11-16</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/">每日一题</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/">动态规划</a></span></div><div class="content">题目题目描述
You are given an integer array $a_1$,$a_2$,…,$a_n$.
The array bb is called to be a subsequence of $a$ if it is possible to remove some elements from $a$ to get $b$.
Array $b_1$,$b_2$,…,$b_k$is called to be good if it is not empty and for every $i$($1\leq i\leq k$) $b_i$ is divisible by $i$.
Find the number of good subsequences in $a$ modulo $10^9$+$7$.
Two subsequences are considered different if index sets of numbers included in them are different. That is, the values of the elements do  ...</div></div></div><div class="recent-post-item"><div class="post_cover right_radius"><a href="/2021/11/15/%E6%AF%8F%E6%97%A5%E4%B8%80DP-3-CF455A-%E7%BA%BF%E6%80%A7DP/" title="每日一DP(3) CF455A (线性DP)">     <img class="post_bg" src="https://pic.imgdb.cn/item/6191f6022ab3f51d9115e383.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="每日一DP(3) CF455A (线性DP)"></a></div><div class="recent-post-info"><a class="article-title" href="/2021/11/15/%E6%AF%8F%E6%97%A5%E4%B8%80DP-3-CF455A-%E7%BA%BF%E6%80%A7DP/" title="每日一DP(3) CF455A (线性DP)">每日一DP(3) CF455A (线性DP)</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2021-11-15T05:52:44.000Z" title="发表于 2021-11-15 13:52:44">2021-11-15</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-11-15T06:35:24.581Z" title="更新于 2021-11-15 14:35:24">2021-11-15</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/">每日一题</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/">动态规划</a></span></div><div class="content">昨天的每日一题因为济南站鸽了一天（虽然是打星）
题目题目描述
Alex doesn’t like boredom. That’s why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence $a$ consisting of $n$ integers. The player can make several steps. In a single step he can choose an element of the sequence (let’s denote it $a_k$) and delete it, at that all elements equal to $a_k + 1$ and $a_k - 1$ also must be deleted from the sequence. That step brings $a_k$ points to the  ...</div></div></div><div class="recent-post-item"><div class="post_cover left_radius"><a href="/2021/11/14/2021ICPC%E6%B5%8E%E5%8D%97%E7%AB%99%E6%B8%B8%E8%AE%B0/" title="2021ICPC济南站游记">     <img class="post_bg" src="https://pic.imgdb.cn/item/6190e7102ab3f51d91c86365.png" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="2021ICPC济南站游记"></a></div><div class="recent-post-info"><a class="article-title" href="/2021/11/14/2021ICPC%E6%B5%8E%E5%8D%97%E7%AB%99%E6%B8%B8%E8%AE%B0/" title="2021ICPC济南站游记">2021ICPC济南站游记</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2021-11-14T10:32:24.000Z" title="发表于 2021-11-14 18:32:24">2021-11-14</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-11-14T11:24:22.630Z" title="更新于 2021-11-14 19:24:22">2021-11-14</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E5%8F%82%E8%B5%9B%E8%AE%B0%E5%BD%95/">参赛记录</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E8%AE%B2%E6%95%85%E4%BA%8B/">讲故事</a></span></div><div class="content">关于这次比赛的流水账先把客户端闪退之类莫名奇妙的事情放一边，就单单说这次比赛。
开局dwg看了一眼G和D，觉得自己之前做过一道类似的题，就开始开D题，下面是D的过题数以及离谱的通过率。

我和glg跟榜开K，一开始想了半天也没理解题目是什么意思，过了30min glg突然说了一句不就是找最短的路径把树给遍历一遍吗，我觉得言之有理，和dwg讨论了讨论就让dwg上机写了。
期间我和glg讨论了一下D感觉应该是三分公差，但是我们两个都没怎么写过三分就卡住了。
20min之后dwg样例没过，一看居然写的bfs，赶紧让glg把dwg替下来，我和dwg一起继续看D。
和dwg说应该是三分，结果dwg也不会写三分，并且咬定自己当初是用差分，前缀和以及后缀和A掉的“此类题”。
glg调bug调到了13点左右，终于过了样例，一交直接wa，全队沉默。
之后又交了一发wa开始自闭。
其实我们的思路一直都有问题，把简单问题复杂化了，我们以为一定是先遍历需要步数最少的，最后遍历步数最多的，导致写了一大堆代码。
中间oms闪退了一次以及又wa了一次K，无伤大雅。
自闭着就来到了封榜，我上机开始码万恶的K。
调bu ...</div></div></div><div class="recent-post-item"><div class="post_cover right_radius"><a href="/2021/11/13/%E6%AF%8F%E6%97%A5%E4%B8%80DP-2-CF191A-%E7%BA%BF%E6%80%A7DP/" title="每日一DP(2) CF191A (线性DP)">     <img class="post_bg" src="https://pic.imgdb.cn/item/618f6bec2ab3f51d914eee77.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="每日一DP(2) CF191A (线性DP)"></a></div><div class="recent-post-info"><a class="article-title" href="/2021/11/13/%E6%AF%8F%E6%97%A5%E4%B8%80DP-2-CF191A-%E7%BA%BF%E6%80%A7DP/" title="每日一DP(2) CF191A (线性DP)">每日一DP(2) CF191A (线性DP)</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">发表于</span><time class="post-meta-date-created" datetime="2021-11-13T07:38:59.000Z" title="发表于 2021-11-13 15:38:59">2021-11-13</time><span class="article-meta__separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-11-13T08:01:49.630Z" title="更新于 2021-11-13 16:01:49">2021-11-13</time></span><span class="article-meta"><span class="article-meta__separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%AF%8F%E6%97%A5%E4%B8%80%E9%A2%98/">每日一题</a></span><span class="article-meta tags"><span class="article-meta__separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/">动态规划</a></span></div><div class="content">题目题目描述
The ancient Berlanders believed that the longer the name, the more important its bearer is. Thus, Berland kings were famous for their long names. But long names are somewhat inconvenient, so the Berlanders started to abbreviate the names of their kings. They called every king by the first letters of its name. Thus, the king, whose name was Victorious Vasily Pupkin, was always called by the berlanders VVP.
In Berland over its long history many dynasties of kings replaced each other, but th ...</div></div></div><nav id="pagination"><div class="pagination"><span class="page-number current">1</span><a class="page-number" href="/page/2/#content-inner">2</a><a class="extend next" rel="next" href="/page/2/#content-inner"><i class="fas fa-chevron-right fa-fw"></i></a></div></nav></div><div class="aside-content" id="aside-content"><div class="card-widget card-info"><div class="is-center"><div class="avatar-img"><img src="https://pic.imgdb.cn/item/6118dff15132923bf87690a5.jpg" onerror="this.onerror=null;this.src='/img/friend_404.gif'" alt="avatar"/></div><div class="author-info__name">FZSF</div><div class="author-info__description">记录与成长</div></div><div class="card-info-data"><div class="card-info-data-item is-center"><a href="/archives/"><div class="headline">文章</div><div class="length-num">20</div></a></div><div class="card-info-data-item is-center"><a href="/tags/"><div 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